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B.Sc II Sem III Chemistry College Name :- Late Ku.Durga K.Banmeru Science College Lonar University Name:- Sant Gadgebaba Amravati University Amravati

- 1. Mr.Shivshankar Purushottam More Assistant Professor Department of Chemisrty Late Ku.Durga K.Banmeru Science College,Lonar Dist.Buldana Captor –VI Liquid State and Electrochemistry
- 2. • Introduction • Matter exists in three states namely solid, liquid and gas. The smallest structural unit of all chemical substances in these three states may be atoms, ions or molecules. The solid exhibits complete ordered arrangement of atoms, ions or molecules while the gaseous state exhibit complete disorder or randomness and the liquid state exhibit only a short-range order. A liquid may be regarded as a condensed gas or a molten solid. In a liquid, the are not rigidly fixed as in solids. They have some freedom of motion which is much more restricted than that in the gases. A liquid, therefore, has a definite volume finite shape. • It is much less compressible and far denser than a gas. Since,the molecule in a liquid are not far apart from one another, the intermolecular forces are • rv strong. The characteristic properties of liquids arise from the nature and the magnitude of these intermolecular forces A.Liquid State
- 3. The surface tension (𝛾) is defined as, the downward force in Newton acting along the surface of a liquid at right angle to any line I meter in length. The unit of surface tension in CGS system is dynes per centimeter (dyne cm"). In SI system, the unit is Newton per meter (Nm-1). Both these units are related as: 1 dyne cm-1 = 10-3 N m-1 • Consider the molecule 'A' in the interior of liquid, which is surrounded from all sides by other molecules. Hence, it is attracted equally in all directions. Therefore molecule 'A’ will behave as if no force is acting on it. However, molecule 'B' which is at the surface of liquid will experience inward pull because it is attracted sideways and towards the interior. A.Surface tension
- 4. • Thus, the surface of liquid tends to contract to the smallest possible area and behaves as if it under tension. This tension which acts along the surface of liquid is called as surface tension. It is the reason that drops of a liquid has spherical shape because for a given volume sphere has minimum surface area due to surface tension • Molecular attraction A.Liquid State
- 5. • In drop number method, the number of drop formed for a fixed volume of liquid is determined by using drop pitter or stalagmometer. The liquid under examination in sucked up in capillary, Say up to upper mark A. then a definite number of drop say 20,are received in weight bottle and weight. form this weight of single drop is calculated. • Let ‘r’ is the radius of capillary tube and 𝛾 is surface tension of liquid. When drop falls off at that time the weight of drop is equal to the force due to the surface tension. • W=2𝜋𝛾---------------(1) Determination of surface tension by drop number method
- 6. For relative determinations, instead of determining the weight of the single drop, number of drops falling between two fixed marks (one above the bulb and other below the bulb) are counted for water (reference). The stalagmometer is dried and filled with another liquid of which surface tension is to be determined. Number of drops for the experimental liquid falling between same two fixed marks is counted. If n, and n2 are the number of drops falling between two fixed marks for water and experimental liquid having densities d1 and d2 and surface tensions liquid falling between same two fixed marks is counted. Determination of surface tension by drop number method
- 7. 𝛾1& 𝛾2 then. When drop falls, wt. of the drop = force due to surface tension W = mg = vdg = 2𝜋r 𝛾-------------(2) .. For water, we can write 2𝜋r 𝛾1= v1d1g--------------(3) For experimental liquid 2𝜋r 𝛾2= v2d2g--------------(4) Multiplying equation (3) and (4) by n, and n, respectively then, 2𝜋r 𝛾1n2=n1 v1d1g----------------(5) 2𝜋r 𝛾1n2 =v2d2g------------------(6) But, n1v1=n2v2=V, the volume of liquid falling between two fixed marks Equations (5) and (6) become Determination of surface tension by drop number method
- 8. 2𝜋r 𝛾1n1=V d1 g---------------(7) 2𝜋r 𝛾2n2=V d2 g---------------(8) Dividing equation (8) by (7), we get 𝛾2 𝛾1 = 𝑛1𝑑2 𝑛2𝑑1 Where , 𝛾2 𝛾1 = 𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑡𝑒𝑛𝑠𝑖𝑜𝑛. Hence by counting number of drop and determining densities, surface tension of any liquid relative to reference (Water) can be calculated. Determination of surface tension by drop number method
- 9. • For all liquid, surface tension (𝛾), decrease with increasing temperature due to increased molecular agitation, these tend to decrease the effect of intermolecular cohesive forces. • At or near the critical temperature, a well defined surface cases to exits and the surface tension become zero. Table: Surface Tension values of some common liquids at various temperature. Effect of temperature on surface tension Liquid Surface tension Nm-1 ×102 273K 293K 313K Water 7.56 7.28 6.96 Banzene 3.16 2.89 2.63 Toulene 3.07 2.84 2.61 Acetone 2.62 2.37 2.12 Ethyl Alcohol 2.40 2.23 2.06
- 10. • i) It is important in the study of emulsion and colloid chemistry. • ii). It is an essential factor in the concentration of ores by froth flotation process. • iii) Surface tension measurements are of much importance in biological science, particularly in bacteriology. • iv) The movement of moisture of soil and passage of sap in plants involve the surface tension. • v)It is used to determine parachor [P]= 𝑀 𝑑 𝛾1/4 a additive and constitutive property used to investigate molecular structures of compounds. • vi) in everyday life, soaps and detergents are used for cleaning purposes. Synthetic surfactants have property of lowering the surface tension of water. Hence, they are used in preparations like tooth paste, cream, toilet soaps, washing powders, medical emulsions, etc. • vii) Surface tension measurements are useful in knowing the presence of air bubble in blood stream and identifying the presence of bile salts in urine (Hay's test for bile salts) Application of Surface Tension
- 11. The surface tension of toluene at 293K is 0.028Nm-1 and its density at this temperature is 0.866×103 if the surface tension of water is 0.07275 Nm-1 and density 0.9982×103 kgm-3m calculate the ration of number of drops of liquid to that of water. Solution: - Given Surface tension of toluene 𝛾 =0.028Nm-1 Surface tension of water 𝛾=0.07275Nm-1 Density of toluene dt=0.866×103kgm-3 Density of water dw=0.9982×103kg-3 Number of drops of toluene =nt Number of drops of water=nw We have 𝛾𝑡 𝛾𝑤 = 𝑛𝑤𝑑𝑡 𝑛𝑡𝑑𝑤 or 𝑛𝑡 𝑛𝑤 = 𝛾𝑤𝑑𝑡 𝛾𝑡𝑑𝑤 𝑛𝑡 𝑛𝑤 = 0.07275×0.866×103 0.028×0.9982×103 =2.25 Problem
- 12. • Every liquid exhibits some resistance to flow. This resistance to flow is called is called viscosity. It is developed in liquid because of the shearing effect of moving one layer of liquid past another. 1 2 3 4 5 Motion of liquid Viscosity of different liquid Viscosity
- 13. • A motion of liquid can be visualized as movement of one layer over another. A layer in contact with stationary surface remains stationary. The second moves slowly, third faster than second and so on. This type of flow is called as laminar flow or streamlines flow. In laminar flow, The force ‘f’ required to maintain a steady difference of velocity dv between two parallel layers separated by distance dx proportional to the area of contact (A) and velocity 𝑑𝑣 𝑑𝑥 • (-) f 𝛼 A. 𝑑𝑣 𝑑𝑥 ----------------(10) • f=ɳ𝐴. 𝑑𝑣 𝑑𝑥 -----------------(11) • Where,f= Retarding force or viscous drag(acting in opposite direction of flow) • ɳ=coefficient of viscosity of the liquid • ɳ = 𝑓 𝐴 𝑑𝑣 𝑑𝑥 ------------------(12) • where, A=1sqm, dv=1msec-1, and dx=1m • then, ɳ= 𝑓 • the coefficient of viscosity (ɳ) may be defined as the force that must be exerted between two parallel layer 1m2 in area and 1 meter apart in order maintain velocity difference of 1 m sec-1
- 14. • In CGS system the unit of ɳ is g cm-1.It is called Poise(P) in particle similar units centipoise (10-2 ) and millipoise (10-3poise) are used. • The SI unit of viscosity is kg m-1 s-1. then are related as • 1 poise =1 g cm-1 =0.1 kg m-1s-1 Units of viscosity:
- 15. • The direct measurement of absolute viscosity of a liquid is very difficult, the relative viscosity of liquid with respect to reference liquid (water) can be conveniently determined with the help of apparatus called Ostwald's Viscometer. • A definite quantity of the liquid under examination is put into the wider limb. The quantity of liquid should be so taken that bigger bulb is filled more than half of its volume (This generally requires 10 to 15 cm-3 of liquid). It is then sucked up into the other limb through a capillary tube. The liquid is allowed to flow through the capillary attached to the smaller bulb and the time of flow from mark A to mark B is noted. Measurement of viscosity by Ostwald’s viscometer method.
- 16. • When liquid flows through the capillary, the time flown ‘ t’ is directly proportional to the viscosity coefficient and inversely to the density ‘d’ of the liquid. 𝑡𝛼 ɳ 𝑑 or ɳ 𝛼 𝑡𝑑 or ɳ = ktd--------(13) The whole process is then repeated for the same volume of water, exactly under similar condition then we have, ɳ𝑤 = 𝑘𝑡𝑤𝑑𝑤-------------------(15) When ɳ𝑤, 𝑘𝑡𝑤 𝑎𝑛𝑑 𝑑𝑤 are the coefficient of viscosity, time of flow and density of water respectively.
- 17. Form the eq.no.(13) and (14) we get, ɳ ɳ𝑤 = 𝑡.𝑑 𝑡𝑤𝑑𝑤 -----------------(15) ɳ ɳ𝑤 =ɳ𝑟 ( i.e the relative viscosity of liquid) Ostwald’s viscometer is a very convenient apparatus for the determination of viscosity at higher temperature as it can be easily suspended in thermostat. Effect of temperature on viscosity It has been found that the viscosity of liquid decrease with rise in temperature. The variation of viscosity with temperature is best expressed by the equation. ɳ=𝐴. 𝑒+𝐸/𝑅𝑇 ------------------(16) Where, A and E (Activation energy for viscous flow) are constant for a given liquid taking logarithms. In ɳ = 𝑖𝑛 𝐴 + 𝐸 𝑅𝑇 -----------------(17) 𝑙𝑜𝑔10ɳ=𝑙𝑜𝑔10A+ 𝐸 2.303𝑅𝑇 -----------------(18)
- 18. Slop= 𝐸 2.303𝑅𝑇 log ɳ 1 𝑇 Hence a plot of log10 1/T should be straight line. the reason why viscosity decrease with temperature is that as that, the temperature increase, the molecule agitations increase and hence the resistance to flow may be expected to decrease.
- 19. 1.Viscosity measurements help in gradation of lubricant oils. In precision instruments such as watches special kinds of lubricants are needed, which should not change their viscosities very much with temperature. Therefore, all weather lubricants are manufactured by mixing long chain coiling polymers with oil. 2.The viscosity measurements yield information regarding the movement of liquid Through Pipes 3. in determination of molecular weight of polymers by viscosity measurements, 4.The study of viscosity has been used by chemist for knowing the constitution of molecules through Parachor R= 𝑑𝑣 𝑑𝑥 ɳ1/8 which is both additive as well as constative properties. 5. When carbon dioxide gets accumulated in blood, breathing becomes difficult. Due to absorption of CO2, blood corpuscles swells which in turn, raise the viscosity of blood. This process quickly lead to heart attack. 6.To the heart patients, doctors prescribe medicines to lower the viscosity of blood in order to lower the pressure on heart. Applications of viscosity measurements:
- 20. Example (2) : Water required 120.5 seconds to flow through a viscometer and the same he volume of acetone required 49.5 seconds. If the densities of water and acetone at 293K are 9.982 x 102 kg m-3 and 7.92 x 102 kg m respectively and the coefficient of viscosity of water at 293 K is 10.05 pascal second, calculate the coefficient of viscosity of acetone at this temperature Solution: Given : Flow time of acetone ta = 49.5 sec. Flow time of water tw = 120.5 sec. Density of acetone da = 7.92 x 10² kg m Density of water dw =9.982 x 10² kg m Coefficient of viscosity of water ɳw= 10.05 Pascal sec. Coefficient of viscosity of acetone ɳa=? We have, ɳ𝑎 ɳ𝑤 = ɳ𝑎𝑑𝑎 𝑡𝑤𝑑𝑤 or ɳa= ɳ𝑎𝑑𝑎 𝑡𝑤𝑑𝑤 × ɳ𝑤 ɳa= 49.5×7.92 ×10 . 05 × 10 2 120 . 5 × 9 . 982 ×102 =3.257 Pascale Sec. Problem
- 21. • Electrochemistry is the branch of physical chemistry which deals with interdependence of chemical charges and electrical energies. Electrochemistry holds a central position in chemistry because it acts as a bridge between thermodynamic and rest of chemistry and it enable to study ionic reaction in detail. • A substance which decomposes on passing current through it is known as electrolyte and phenomenon of decomposition is called electrolysis. • Spontaneous oxidation-reduction (redox) reaction can be used to produce electric current under suitable condition. The device used for this purpose is called electrochemical cell or Galvanic cell Chemical reaction can be forced to proceed by passing electric current. A device used for this purpose is called electrolytic cell. B. Electrochemistry
- 22. The capacity of conductor to carry the electrical current (Energy) is known as the conductance or conductivity. We generally come across with two conductors. i.e metallic and electrolytic conductors. 𝐶 1 𝑅 It is measured in ohm-1 or mho or siemens (S) Resistance of any uniform conductor varies directly as its length and inversely to its area of cross section. 𝑅 𝛼 𝑙 𝑎 = 𝜌. 𝑙 𝑎 Where 𝜌 = constant called specific resistance or resistivity l= length of conductor a= cross section area of conductor Conductance of electrolyte solution.
- 23. When, l=a=1, R=ρ Hence, Specific resistance is the resistance of conductor of unit length and unit cross sectional area or it is resistance of 1m3 material. Specific conductance (k): “ The conductance of one centimeter cube (1cm3) or one cubic meter(1m3) solution of an electrolyte is known as specific conductance.” It is denoted by ‘k’ (Kappa).Specific conductance (k) is reciprocal of specific conductance(𝜌). ∴ 𝑘 = 1 𝜌 𝑜𝑟 𝑘 = 𝐶. 𝑙 𝑎 𝑂𝑅 𝑘 = 1 𝑅 . 𝑙 𝑎 Where, C = conductance,R=Resistance ,l= length of conductance & a= cross section area of conductor. For electrolyte specific conductance is the conductance of one meter cube of solution. Its CGS unit is ohm-1 or S cm-1 & SI unit is ohm-1 m-1 or S m-1
- 24. • Conductance of an electrolyte solution is the reciprocal of resistance (C-1/R) Therefore, measurement of conductance is done indirectly by determining the resistance of the solution. The resistance is measured by Wheatstone AC bridge method. Direct current cannot be used in process as this will give wrong result due to 1)Change in concentration due to electrolyte 2)Change in resistance due to polarization at electrode. Determination of conductance of electrolyte solution
- 25. • These difficulties are overcome by using alternating current within audio frequencies range and galvanometer in Wheatstone bridge is replaced by headphone. The schematic diagram of the apparatus is shown in fig. • The solution whose conductivity is to be determine is taken in a suitable conductivity cell ‘C’. When current is flowing know resistance ‘R’ is introduced through resistance box. The sliding contact ‘X’ is then moved along the wire ‘AB’ of uniform thickness until a point of minimum sound is detected in headphone (G). At this stage, 𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑜𝑥 = 𝐿𝑒𝑛𝑔𝑡ℎ 𝐵𝑋 𝐿𝑒𝑛𝑔𝑡ℎ 𝐴𝑋
- 26. As all values from above equation are known, resistance of solution can be determined. Form this resistance, Conductance of solution can be determined. Determination of cell constant. Cell constant (𝜃) is defined as the ratio of length( distance) between the electrode ‘l’ and area of cross section ‘a’ of electrode. ∴ Cell constant (𝜃)= 𝑙 𝑎 In order to determine cell constant (𝜃) it is necessary to determine ‘l’ & ‘d’ but actually it is not possible. So indirect method based on measurement of conductance of standard KCl solution is employed as flows: Specific conductance = 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒 𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒 to determine the cell constant, a standard solution of KCl of known specific conductance at a given temperature is used. Its conductance is determined experimentally at the same temperature. Substituting the two values is above equation, the cell constant can be calculated.
- 27. A Conductivity cell was filled with 0.1M KCl which was known to have specific conductance of 0.1404 mhom-1 at 298K.Its measure resistance at 298K was 99.3 ohm. When the cell was filled with 0.02M AgNO3m,the resistance was 50.3 ohm. Calculate i) Cell Constant,(ii) specific conductance of AgNO3 solution. Solution: - For 0.01M KCl K= 0.1404mho-1 R= 99.3 ohm Specific conductance (k) =cell constant × Observed conductance ∴ Cell Constant= 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒 𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝐶𝑜𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒 = 0.1404 1/99.3 =0.1404 ×99.3 =13.94 m-1 ii)For 0.02 M AgNO3, R =50.3 ohm Sp. Conductance =Cell Constant × 1 𝑅 =13.94× 1 50.3 =0.2771 mho-1 Problem
- 28. Example (2): 0.5 N solution of salt occupying volume between two platinum electrodes 0.0172 m apart and 0.04499 sq. m. area has resistance 25 ohm. Calculate equivalent conductance of solution. Solution: Given data, Distance between electrodes, 1 = 0.0172 m area of cross section of electrode a = 0.04499 sq. m, Resistance of salt solution R = 25 ohm concentration of salt solution = 0.5N Sp. conductance (k) = Cell constant x Observed conductance = 0.0172 0.04499 × 1 25 =0.3823 × 0.04 = 0.01529 mho-1 Eq. Conductance 𝜆𝑣= Sp. Conductance × V 𝜆𝑣=k× 𝑉 = 𝑘×1000 𝑁 = 0.01529×1000 0.5 =38.58 S m2 equv-1
- 29. Example (3): The resistance of conductivity cell was 7.02 ohm when filled with 0.1 N KCl solution (k=0.1480 ohm m') and 69.2 ohm, when filled with N/100 NaCl solution at same temperature. Calculate the cell constant and equivalent conductance of NaCl Solution: (1) For 0.1 N KCI, R= 7.02 ohm, k = 0.1480 ohm’m Cell constantCell Constant = Specific conductance 𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝐶𝑜𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒 0.1480 1/7.02 =0.1480 × 7.02 = 1.038 m-1 (ii) For N/100 or 0.01 N NaCl, R = 69.2 ohm Sp. Conductance k = Cell constant x Observed conductance 1.038 × 1 69.2 = 0.015 𝑆 m-1 Equivalent conductance = Sp. conductance x V 𝜆𝑣=k× 𝑉 = 𝑘×1000 𝑁 = 0.015×1000 0.01 = 1500 S m² equiv-1
- 30. • Conductance of an electrolyte solution is the reciprocal of resistance (C-1/R) Therefore, measurement of conductance is done indirectly by determining the resistance of the solution. The resistance is measured by Wheatstone AC bridge method. Direct current cannot be used in process as this will give wrong result due to 1)Change in concentration due to electrolyte • 2)Change in resistance due to polarization at electrode. Determination of conductance of electrolyte solution
- 31. These difficulties are overcome by using alternating current within audio frequencies range and galvanometer in Wheatstone bridge is replaced by headphone. The schematic diagram of the apparatus is shown in fig. The solution whose conductivity is to be determine is taken in a suitable conductivity cell ‘C’. When current is flowing know resistance ‘R’ is introduced through resistance box. The sliding contact ‘X’ is then moved along the wire ‘AB’ of uniform thickness until a point of minimum sound is detected in headphone (G). At this stage, 𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑜𝑥 = 𝐿𝑒𝑛𝑔𝑡ℎ 𝐵𝑋 𝐿𝑒𝑛𝑔𝑡ℎ 𝐴𝑋 As all values from above equation are known, resistance of solution can be determined. Form this resistance, Conductance of solution can be determined.
- 32. Cell constant (𝜃) is defined as the ratio of length( distance) between the electrode ‘l’ and area of cross section ‘a’ of electrode. ∴ Cell constant (𝜃)= 𝑙 𝑎 In order to determine cell constant (𝜃) it is necessary to determine ‘l’ & ‘d’ but actually it is not possible. So indirect method based on measurement of conductance of standard KCl solution is employed as flows: Specific conductance = 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒 𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒 to determine the cell constant, a standard solution of KCl of known specific conductance at a given temperature is used. Its conductance is determined experimentally at the same temperature. Substituting the two values is above equation, the cell constant can be calculated. Determination of cell constant.
- 33. A Conductivity cell was filled with 0.1M KCl which was known to have specific conductance of 0.1404 mhom-1 at 298K.Its measure resistance at 298K was 99.3 ohm. When the cell was filled with 0.02M AgNO3m,the resistance was 50.3 ohm. Calculate i) Cell Constant,(ii) specific conductance of AgNO3 solution. Solution: - For 0.01M KCl K= 0.1404mho-1 R= 99.3 ohm i) Specific conductance (k) =cell constant × Observed conductance ∴ Cell Constant= 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒 𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝐶𝑜𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒 = 0.1404 1/99.3 =0.1404 ×99.3 =13.94 m-1 ii)For 0.02 M AgNO3, R =50.3 ohm Sp.Conductance =Cell Constant × 1 𝑅 =13.94× 1 50.3 =0.2771 mho-1 Problem
- 34. Example (2): 0.5 N solution of salt occupying volume between two platinum electrodes 0.0172 m apart and 0.04499 sq. m. area has resistance 25 ohm. Calculate equivalent conductance of solution. Solution: Given data distance between electrodes, 1 = 0.0172 m area of cross section of electrode a = 0.04499 sq. m. resistance of salt solution R = 25 ohm concentration of salt solution = 0.5N Sp. conductance (k) = Cell constant x Observed conductance = 0.0172 0.04499 × 1 25 =0.3823 × 0.04 = 0.01529 mho-1 Eq.Conductance 𝜆𝑣= Sp.Conductance × V 𝜆𝑣=k× 𝑉 = 𝑘×1000 𝑁 = 0.01529×1000 0.5 =38.58 S m2 equv-1
- 35. Example (3): The resistance of conductivity cell was 7.02 ohm when filled with 0.1 N KCl solution (k=0.1480 ohm m') and 69.2 ohm, when filled with N/100 NaCl solution at same temperature. Calculate the cell constant and equivalent conductance of NaCl Solution: (1) For 0.1 N KCI, R= 7.02 ohm, k = 0.1480 ohm'm Cell constant Cell Constant = Specific conductance 𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝐶𝑜𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒 0.1480 1/7.02 =0.1480 × 7.02 = 1.038 m-1 (ii) For N/100 or 0.01 N NaCl, R = 69.2 ohm Sp. Conductance k = Cell constant x Observed conductance 1.038 × 1 69.2 = 0.015 𝑆 m-1 Equivalent conductance = Sp. conductance x V 𝜆𝑣=k× 𝑉 = 𝑘×1000 𝑁 = 0.015×1000 0.01 = 1500 S m² equiv-1
- 36. The solution of electrolyte conducts electricity due to the presence of ions. The conductivity at constant temperature is approximately proportional to number of ions. According to ionic theory the number of ions increases as the solution of the electrolyte is progressively diluted. Therefore, equivalent conductivity increases as the dilution increases. The specific conductivity decreases on dilution because on dilution the number of ions per dm3 solution decreases in spite of increase in dissociation. The equivalent and molecular conductivity increases because these are the product of specific conductivity and volume containing on gram equivalent or one mole of an electrolyte. 𝜆𝑣= k. V On dilution, decrease in specific conductance is compensated by increase in volume 'V'. Variation of specific and equivalent conductance with dilution
- 37. • Table: Specific and equivalent conductance of KCI solutions at different concentrations (dilutions) at 18°C • It is seen that equivalent conductance increases with dilution. The increase in case of electrolytes like HCI, KCl and BaCl2, is not so large as in case of acetic acid or NH4OH. The electrolytes of first category (HCI, KCl, BaCl2, etc.) are known as strong electrolytes while in those of second category (CH3COOH, NH4OH, etc.) are known as weak electrolytes. This behavior can be illustrated from the table. Concentration (Kg equiv.per dm3) Specific conductance ×10-4 (Sm-1) Equivalent conductance ×10-4 (Sm2equiv-1 1.00 0.0982 98.2 0.10 0.01120 112.0 0.01 0.001223 122.3 0.001 0.0001273 127.3 0.0001 0.00001291 129.1
- 38. • In case of strong electrolyte the equivalent and molecular conductivity progressively increase on dilution and rise to maximum limit. This limiting value of conductivity is describe as equivalent conductivity at infinite dilution (𝜆∞) • In case of weak electrolyte conductivity continues to increase with dilution for all practical values of dilution.The maximum is reached when dilution is infinite. Therefore,for weak electrolyte,There is no indication that limiting value can be attained even when concentration approaches to zero.
- 39. • The titration in which end point is determine by measuring change in conductance of solution upon addition of reagent is called conductometric titration. The conductance of a solution depends largely on the number of ions and their mobilities. Some examples of conductometric titrations are as given below. • 1) Strong acid against strong base: (HCI against NaOH) Consider the titration of strong acid (HCI) against strong base (NaOH). The acid is taken in conductivity vessel and alkali in burette. The conductance of HCl is due to presence of H and Cl ions. As alkali is added, gradually H ions are replaced by slow moving Nat ions as given below: H+ + CI-+ Na+ + OH- → Na+ +CI-+H2O Until the complete neutralization, conductance decreases on addition of NaOH. Any subsequent addition of alkali after end point will result in introduction of fast moving OH ions. The conductance therefore increases on further addition of alkali. Conductometric titrations
- 40. • The variation of conductance is plotted against volume of alkali added, we get two straight lines meeting at point 'B' which represents end point of titration.
- 41. • When acid is weak, conductance is low, on addition of strong base poorly conducting acid is convert into highly ionsed salt and hence conductance increase slowly up to the equivalence point. Beyond the equivalence point addition of alkali causes sharp increase in conductance due to excess of hydroxide ions. The graph is represented as: 2)Weak acid against strong Base (CH3COOH against NaOH)
- 42. • In this case conductance initially decrease due to the replacement of fast moving H+ ions by slowly moving NH4+ ions. Beyond end point, further addition of weakly ionized HN4OH will not cause any appreciable change in conductance. The point of intersection of curve is the end point of titration. 3)Strong acid against weak base:(HCl against NH4OH)
- 43. • In this titration, conductance initially increases because of formation of salt (CH3COONH4) which is strong electrolyte. This increase continues till end point. Beyond end point, the conductance does not change appreciable. The graph is shown in fig. 4)Weak acid against weak base: (CH3COOH against NH4OH)
- 44. The iteration of silver nitrate against potassium chloride involves precipitation formation. AgNO3 + KCl KNO3 + AgNO3 - Since mobility of Ag+ and K+ ions is nearly same, the conductance remains almost constant till the equivalence point. After equivalence point the added KCl, increases The conductance rapidly as shown in graph given below. 5)Precipitation titration: (AgNO3 against KCl)
- 45. The conductometric titration has many advantages over ordinary titration: i) Small quantity of solutions is required for titrations. ii) As end point is determined graphically, no special precautions are necessary iii)Indicator is not required, so conductometric titrations are used in titration of colored and turbid solutions. iv)Conductometric titrations are used for analysis of dilute solutions as well as for weak acids. v) Conductometric titrations can be applied to mixture of acids, precipitation & other types of titrations, vi) Conductometric measurements give more accurate results. Advantages of conductometric titration:
- 46. The conductometric titration has many advantages over ordinary titration: i) Small quantity of solutions is required for titrations. ii) As end point is determined graphically, no special precautions are necessary iii)Indicator is not required, so conductometric titrations are used in titration of colored and turbid solutions. iv)Conductometric titrations are used for analysis of dilute solutions as well as for weak acids. v) Conductometric titrations can be applied to mixture of acids, precipitation & other types of titrations, vi) Conductometric measurements give more accurate results. • Advantages of conductometric titration:
- 47. Migration of ions under influence of electric field • On passing electric current through electrolyte solution, ions migrate and discharged oppositely charged electrodes. The migration of ions can be demonstrated by simple experiment. • The lower portion of U-tube is filled with 5% agar-agar solution in water with small quantity of CuCr2O- (obtained by mixing equimolar quantities of K2Cr2O, + CuSO4). It is allowed to set by cooling as dark green jelly. Some charcoal powder is sprinkled in both limbs. Then solution of KNO3, and agar-agar is placed in each limb and allowed to set as jelly.
- 48. • Finally, solution of KNO; in water is filled in each limb and platinum electrodes are placed as shown in figure • When electric current is passed, Cu+ ions migrate towards cathode (-ve electrode).Due to this blue colour appears in cathode side and yellow color in anode side by Cr207 -2. ions. From the movement of these colour bands, speed of ions can be compared.
- 49. • Although most of ions differ in their mobilities, the total number of ion discharged at electrodes on electrolysis is same. This can be explained by Hittorf's theoretical device as shown in figure. It consists of an electrolytic cell containing same number of positive and negative ions with same valency. The electrolytic cell is divided into three compartments by porous partitions B & C. Metal electrodes A and D represent cathode and anode respectively. Hittorf's theoretical device
- 50. i) Represent initial state of electrolyte solution before electrolysis in which equal number of positive and negative ions are present. On passing electric current, three cases may be arise: - i)suppose, only anions are migration and cations remains stationary. If speed of anion (v=0) and speed on cation (u=0).then two anions migrate form cathode to anode compartment and are discharge at anode. ii)The unpaired cation in cathode compartment are discharge at cathods.Therefore the number of ions discharge in both compartment is two. iii)suppose,both ions are moving with same speed. Let, u=v=2,then the number of ions discharge at respective electrode is four. iv)suppose, both ions are migrating with different speeds (u=1,v=2).in this case total number of ions discharge is three.
- 51. i) During electrolysis, ions are discharge in equivalent amount, Irrespective of their speeds of migration. ii) Concentration of electrolyte, around changes as a result of migration of ions. iii)Fall in concentration around electrode is directly proportional to the speed of that ion which moves away from that electrode. ∴ 𝐹𝑎𝑙𝑙 𝑖𝑛 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑎𝑟𝑜𝑢𝑛𝑑 𝑎𝑛𝑜𝑑𝑒 𝐹𝑎𝑙𝑙 𝑖𝑛 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑎𝑟𝑜𝑢𝑛𝑑 𝑐𝑎𝑡ℎ𝑜𝑑 = 𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝑐𝑎𝑡𝑖𝑜𝑛 𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑛𝑖𝑜𝑛 𝑢 𝑣 = 𝑟(𝑠𝑝𝑒𝑒𝑑 𝑟𝑎𝑡𝑖𝑜) iv) Total current carries by solution is measure of (u+v).This holds good when the electrodes or ions are not attacked by solution. In the above illustration concentration of central compartment remains constant. Whatever be the speed of ions,the number of ions discharge on electrode is always equal. Following conclusion can be drawn about the process of electrolysis:
- 52. During electrolysis current is carried by cation and anions. “The fraction of the total current carried by an ionic species is called its transference number or transport number.The transport number of cation(𝑡+) and anion (𝑡−) is given as. 𝑡+= 𝑢 𝑢+𝑣 𝑡−= 𝑣 𝑢+𝑣 u=Speed of cation v= speed of anion “Sum of two transport number will be one” 𝑡+ + 𝑡− = 1 𝑜𝑟 1 − 𝑡+ If the speed ratio is r= 𝑢 𝑣 = 𝑡+ 𝑡− ∴ r= 𝑡+ 𝑡− = 𝑡+ 1−𝑡+ = 1−𝑡− 𝑡− ∴r𝑡−=1-𝑡− or 𝑡−+r𝑡−=1 or 𝑡−(1+r)=1 and 𝑡−= 1 1+𝑟 Transference number of transport number of Hittrof’s number of ions.
- 53. Problem(3) The speed ratio of silver and nitrate ions in a solution of silver nitrate electrolysed between silver electrode is 0.916.find the transport number of the two ion. Solution. We have 𝑡−= 1 1+𝑟 Where 𝑡− is the transport number of anion and r is the speed ratio of cation and anion. ∴ 𝑡𝑁𝑂− 3 = 1 1+0.916 =0.521 and 𝑡𝐴𝑔+=1-𝑡𝑁𝑂− 3 =1-0.521=0.479
- 54. • This method is based on the fact that change in concentration around the electrodes is due to migration of ions. This apparatus used in this method is shown in figure .It consist of two vertical glass tubes connected through U-tube. All the three tubes are provided with stop cock at bottom. Cathode is a portion of it is exposed to solution in the form of spiral. . The apparatus is connected to copper or silver voltmeter in series. The apparatus is filled with standard solution of silver nitrate. 1.Hittrof’s Method
- 55. A steady current of 0.01 ampere is passed for nearly two to three hours. At the end of this period suitable quantity of solution is drawn form lower portion of anode limb and its weight is determined. It is then titrated with standard solution of potassium thiocyanate solution to determine amount of silver present in it. The weight of silver deposited in voltmeter is noted. If copper voltmeter us is used the weight of copper deposited is multiplied by 108/31.05 to determined silver equivalent. Precautions: i) Steady current should be passed. ii) There should be no change in concentration of middle compartment. Observation and calculation: Since, the change in concentration is accompanied by change in volume, loss of material must be determined with reference to definite weight of solvent present after current has passed. Two different cases may arise:
- 56. Case I. When electrodes are non-attackable (Pt electrodes are used): . After passing electric current: Let the weight of anodic solution taken out = a gm,Weight of AgNO3 present in it by titration=b gm Weight of water = (a - b) gm Before passing electric current: Let the weight of AgNO3 in (a - b) gm of water before passing electric Current = c gm ... Fall in concentration (c-b) gm of AgNO3 = 𝑐−𝑑 170 gm equvt.of AgNO3 = d (Say) Let,the weight of silver deposited in silver coulometer= w1 gm 𝑤1 108 gm equvt of Ag W= (say) gm equvt. of Ag ∴=Transport number of 𝐴𝑔+,𝑡+= 𝐹𝑎𝑙𝑙 𝑖𝑛 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑎𝑟𝑜𝑢𝑛𝑑 𝑎𝑛𝑜𝑑𝑒 𝐴𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝐴𝑔 𝑑𝑒𝑝𝑜𝑠𝑖𝑡𝑒𝑑 𝑖𝑛 𝑔𝑚 𝑒𝑞𝑢𝑖𝑣𝑡. = 𝑑 𝑊 and transport number of NO3-,𝑡−= 1- 𝑑 𝑊 = 𝑊−𝑑 𝑊
- 57. Case II When electrode are attackable (Ag electrode are used) In this case b>a because NO3 - ions react with Ag anode to produce AgNO3.Thus concentration of AgNO3 or Ag is increased in anode compartment. Increase in conc. of anodic solution=(b-c) gm of AgNO3 = 𝑏−𝐶 170 gm equvt of AgNO3 = e (say) If no Ag+ ions had migration from the anode, the increase in concentration of Ag+ ions would have been equal to W ∴ Fall in concentration due to migration of Ag+ ion =(W-e) Hence, Transport number of Ag+.𝑡+ = 𝑊−𝑒 𝑊 And Transport number of NO3- 𝑡−=1- 𝑊−𝑒 𝑊
- 58. Hittorf's method does not give accurate results for dilute solutions. Also small changes in concentration due to passage of electric current may lead to experimental error. Hence, direct observation method called "Moving Boundary Method" was devised by Lodge and improved by Whetham. i)There are three conditions for determination of transport number, ii) Cation of indicator electrolyte should not move faster than the cation whose transport number is to be determined. ii) Both have same anion iii) Indicator electrolyte should have more density. Moving Boundary Method:
- 59. • It consists of electrolytic cell with vertical tube of uniform cross section with two electrodes at two ends. At lower side anode is made of cadmium rod and cathode is Pt foil. • If we have to determine transport number of cation H+ in HCl, we have to choose another electrolyte called "Indicator electrolyte' having common anion with the chemical species under study (here Cl -ion). The cation of indicator electrode must be slow moving as compared with the cation whose transport number is to be determined. Here serves as indicator electrode because cd+ ion moves slowly than H' ion. This prevents blurring of boundary line. The indicator solution CaCd2, is placed in lower half and over this the solution of HCl is allowed to float, to produce sharp boundary at B, between two solution the cadmium chloride produced during electrolysis will itself act as indicator electrolyte. • A constant current is passed through apparatus for 5 to 6 hours. The H+ ion move towards cathode, followed by Cd+2ions. The boundary gradually moves upwards up to B2, i.e. through a distance T m.
- 60. Hence, quantity of current carried by H' ions = t. Q ; Hence, amount of H' ions migrated from B, to B2 = 𝑄.𝑡+ 𝐹 kg equivalent--------------(1) If 'x' sq. m. is cross sectional area of tube, then volume between B1, to B2, will be = x. l. dm3 If 'c' is the concentration of H+ ions in kg equivalent per liter (dm3) then amount of H+ ions in given volume. = 𝑥.𝑙.𝑐 1000 kg equivalent------------------(2) From eq.(1) Ana (2) 𝑄.𝑡+ 𝐹 = 𝑥.𝑙.𝑐 1000 Or 𝑡+= 𝑥.𝑙.𝑐.𝐹 1000.𝑄 If 𝑛 𝑄 𝐹 ∴ 𝑡+= 𝑥.𝑙.𝑐. 1000.𝑛 And 𝑡−= 1-𝑡+ Where Q= Quantity of electricity passed= ampere x second n= number Faradays of current passed 𝑡+= Transport number of H+ ion (1F= 965000 coulomb)
- 61. Calculate the transport number of H+ and Cl-ions tor from electrode. Concentration of HCI solution c = 0.100 N. Mass of the silver deposited in coulometer = 0.1209g Distance up to which boundary moves = 1.24 cmArea of cross section of tube = E 7.5 cm. Solution: Transport number of H+ ion 𝑡+ 𝑡+= 𝑥.𝑙.𝑐.𝐹 1000.𝑄 = 𝑡+= 𝑥.𝑙.𝑐. 1000.𝑛 = l=7.5m; x=1.24m2;c=0.1N x= 1.24 m; c= 0.1 N l = 7.5 m Mass of silver in coulometer=0.1209g Number of Faradays of current passed in circuit 'n' can be calculated as 108 g of silver = 1 Faraday (F) 0.1209 g of silver= 0.1209 108 F i.e. n= 7.5×1.24×0.1×108 1000×0.1209 =0.8308 Transport number of Cl- ion 𝑡−=1-0.8308 =0.1692 Problem
- 62. • Problem A solution of LiCl of molarity 0.10 was placed in moving boundary cell cross-section area 1.17 m and was electrolyzed for 131 minutes with a constant current of 9.42 x 10-3 A. The Li+ boundary was observed to move a distance of 2.08 m What is the transference number of Li+ ions in this solution? • Given data Area of cross section of tube ‘x’=1.17m2 Distance up to which boundary moves l = 2.08 m. Concentration of LiCl solution 'c' = 0.1 mol dm-3 Current passed = 9.42 x 10-3A Time 't' = 131 min = 131 x 60 = 7860 sec. Q = current x time=9.42 x 10-3 x 7860 Transport number of Li+ 𝑡 +𝐿𝑖= 𝑥.𝑙.𝑐.𝐹 1000.𝑄 = 1.17×2.08×0.1×96500 1000×9.427×10−3×7860 𝑡 +𝐿𝑖=0.317 electrolysis between Platinum electrodes 100 dm cathode solution contained 0.06315
- 63. Problem The original strength of NaOH solution was 0.059 kg per 100 dm3. After Ag of NaOH and at same time current deposited 0.05216 kg silver in voltmeter. Find out transport number of Na and OH ions. Solution: Given data Conc, of cathode solution before electrolysis = 0.059 kg NaOH/100 dm Conc, of cathode solution after electrolysis = 0.0631 kg NaOH/100 dm' Actual increase in conc, = (0,06315) -(0.059) = 0.00415 kg NaOH If no hydroxide ion had migrated from cathode then, Theoretical increase = Total deposition in voltmeter = 0.05216 kg Ag = 0.05216 x 0.040/0.108 kg NaOH = 0.01932 kg NaOH Hence, decrease in conc. around cathode = theoretical increase - actual increase = 0.01932 -0.00415= 0.01517 ∴ Transport number of OH-ion 𝑡 −𝑂𝐻 = 𝐷𝑒𝑐𝑟𝑒𝑎𝑠𝑒 𝑐𝑜𝑛.𝑎𝑟𝑜𝑢𝑛𝑑 𝑐𝑎𝑡ℎ𝑜𝑑 𝑇𝑜𝑡𝑎𝑙 𝑑𝑒𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑖𝑛 𝑣𝑜𝑙𝑡𝑚𝑒𝑡𝑒𝑟 𝑡 −𝑂𝐻= 0.01517 0.01932 =0.785 𝑡 +𝑁𝑎=(𝑡 −𝑂𝐻) = (1-0.785)=0.215
- 64. Example (8) The speed ratio of silver and nitrate ions in a solution of AgNO3, electrolyzed between two electrodes is 0.9. Find the transport number of Ag+ and NO3 - Solution: Transport number of (NO3 -) 𝑡−= 1 1+𝑟 = 1 1+0.9 =0.5262 Transport number of (Ag+) 𝑡+= (1-𝑡−) = (1 -0.5262) = 0.4738
- 65. Kohlrausch (1875) observed that at infinite dilution, where ionization of electrolyte is complete, each ion migrates independently and contributes to total equivalent conductance of 1, & ne is the ionic conductance of anion and cation respectively. Kohlrausch electrolyte. He has given a generalization called Kohlraush's law. It states that, “the equivalent conductance of an electrolyte at infinite dilution is the sum of equivalent conductance of anion and cation." i.e. 𝜆∞= 𝜆𝑎 + 𝜆𝑐 • 𝜆𝑎& 𝜆𝑐 is the ionic conductance of ions and cation respectively. Kohlrausch studies studied equivalent conductance at infinite dilution for various electrolyte having same cation or anion which can be given in table. Kohlrausch's law of independent migration of ions
- 66. Table: Equivalent conductance at infinite dilution at 18°C It is evident from the values that difference in conductance of any two cations appears to be same irrespective of nature of anion with which it is associated. Similarly, difference in conductance of any two anions appears to be same irrespective of nature of cations with which it is associated. Pair of Electrolyte 𝜆∞ × 10−4 𝑆𝑚2 𝑒𝑞𝑢𝑖−1 Difference × 10−4 𝑆𝑚2 𝑒𝑞𝑢𝑖−1 KCl NaCl 130.0 108.9 21.1 KNO3 NaNO3 126.3 105.2 21.1 KCl KNO3 130.0 126.3 3.7 NaCl NaNO3 108.9 105.2 3.7
- 67. The ionic conductance's are proportional to velocities of ions. 𝜆𝑎 𝛼 𝑣 or 𝜆𝑎=k.v -----------(1) & 𝜆𝑐 ∝ u or 𝜆𝑐=k.v--------------(2) ∴ 𝜆𝑎 + 𝜆𝑐 = k(v+u)------------(3) Form Kohlrauch’s law 𝜆𝑎 + 𝜆𝑐 = 𝜆∞ 𝜆∞= k(u+v)-----------(4) Dividing equation (1) by (3) 𝜆𝑎 𝜆∞ = 𝑣 𝑢+𝑣 =𝑡− ∴ 𝜆𝑎=𝑡−.𝜆∞ & 𝜆𝑎 𝜆∞ = 𝑣 𝑢+𝑣 =𝑡− ∴ 𝜆𝑎=𝑡−.𝜆∞ −−−−− −(5) & 𝜆𝑐 𝜆∞ = 𝑢 𝑢+𝑣 =𝑡− ∴ 𝜆𝑐=(1 − 𝑡−).𝜆∞ OR ∴ 𝜆𝑐=𝑡+.𝜆∞--------(6) Thus,the ionic conductance of any is the product of its own transport number and 𝜆∞value for any strong electrolyte containing that ion. Its is express as mho m2, equiv-1 Relation between transport number & ionic conductance:
- 68. Weak electrolytes do not ionize to a sufficient extent in solution and are not beingcompletely ionized even at very great dilution. The practical determination of equivalent conductance at infinite dilution '2' in such cases is therefore, not possible. However, it can calculate with the help of Kohlrausch's law, e.g. the equivalent conductance at infinite flation of CH3COOH (weak electrolyte) can be obtained from the equivalent conductance at infinite dilution of HCI, CH3COONa and NaCl (all of which are strong electrolytes) as given below, Now i)𝜆∞(HCl) = 𝜆∞(H+) + 𝜆∞(Cl-) = 425.0 S m2 equvi-1 ii) 𝜆∞(CH3COONa)= 𝜆∞(Na+) + 𝜆∞ (CH3COO-) 91.6 S m2 equvi-1 iii) 𝜆∞(NaCl)= + 𝜆∞(Na+) + (Cl-)=128.1 S m2 equvi-1 From eq.(i),(ii) and (iii) 𝜆∞(H+)+ (CH3COO-)=𝜆∞(H+)+𝜆∞(Cl-)+𝜆∞(Na+)+(CH3COO-)-𝜆∞(Na+)+𝜆∞(Cl-) 𝜆∞(Ch3COOH)= 𝜆∞(HCl)+ 𝜆∞(CH3COONa)- 𝜆∞(NaCl) 𝜆∞(Ch3COOH)= 425.0+ 91.6-128.1= 388.5 S m2 equvi-1 In this manner, the equivalent conductance at infinite dilution of weak electrolyte can be calculated from the equivalent conductance at infinite dilution of strong electrolytes containing similar ions. Applications of conductivity measurement
- 69. 2. Determination of degree of dissociation (a) of weak electrolyte: Degree of dissociation (𝛼) of a weak electrolyte at any dilution can be calculated by the relationship 𝛼 = 𝜆𝑐 𝜆∞ Where, 𝜆𝑐 = equivalent conductance at given concentration 'c' 𝜆∞ = equivalent conductance at infinite dilution. The dissociation constant (𝑘𝑎) is defined as, "equilibrium constant for dissociation of m electrolyte obtained by applying law of mass action at given temperature." eg. Consider dissociation of a weak acid HA HA H+ + A-
- 70. For strong electrolyte ka has higher value i.e value ka indicates grater degree of dissociation (𝛼) can be given as HA H+ + A- C 𝛼 𝛼 (1- 𝛼) 𝛼. 𝐶 𝛼. 𝐶 𝑘𝑎= 𝐻+ 𝐴− 𝐻𝐴 = (𝛼.𝑐)(𝛼.𝑐) 1−𝛼 .𝑐 ∴ 𝑘𝑎= 𝛼2.𝑐 1−𝛼 ----------(1) But, in case of weak electrolyte ,ka has lower value i.e degree of dissociation (𝛼) is negligible, hence ,the dissociation constant of weak electrolyte can be given can be given as ∴ 𝑘𝑎= 𝛼2 .c From the value of equivalent conductance at given concentration (𝜆𝑐) and equivalent conductance at infinite dilution (𝜆∞) degree of dissociation (𝛼) cab be determine. 𝛼= 𝜆𝑐 𝜆∞ Hence, form conductivity measurement dissociation constant Ka for weak electrolyte can be determine.
- 71. Example (9): At 25°C the transport number of H+ ions In HCl and CH3COO- lon in CH3COONa are 0.81 and 0.47 respectively. The equivalent conductance at Infinite dilution of HCl and CH3COONa are 426 ohm.-1cm2 equvt-1 and 91.0ohm-1 cm2 equvt-1 respectively. Calculate the equivalent conductance of acetic acid at infinite dilution. Solution: Conductance of H+ ion; Ac (H.T) = t. (H) x 200(HCI) = 0.81 x 426 = 345.06 Conductance of CH3COO- ion; 1(CH3COO-) = t. (CH3COO-) x (CH3COONa) = 0.47 x 91 = 42.77 Conductance of acetic acid at infinite dilution; 2 (CH3COOH) = 2, (CH3C00) + Rc (H) = 42.77 + 345.06 = 387.83 mhos Problem